Answer
$$ e^{-t} \cos (2t)-\dfrac{1}{2} e^{-t}\sin (2t) $$
Work Step by Step
Since, $F(s)=\dfrac{s}{(s+1)^2+4}=\dfrac{s+1}{(s+1)^2+2^2}-\dfrac{1}{(s+1)^2+2^2}$
The inverse Laplace transform of function can be expressed as:
$F(t)=L^{-1} [\dfrac{s+1}{(s+1)^2+2^2}-\dfrac{1}{(s+1)^2+2^2}]$
Now, apply the first shifting Theorem.
$f(t)= e^{-t} \cos (2t)-\dfrac{1}{2} e^{-t}\sin (2t) $
